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150k^2-216=0
a = 150; b = 0; c = -216;
Δ = b2-4ac
Δ = 02-4·150·(-216)
Δ = 129600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{129600}=360$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-360}{2*150}=\frac{-360}{300} =-1+1/5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+360}{2*150}=\frac{360}{300} =1+1/5 $
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